Simplify and expand the following expression: $ \dfrac{t}{t - 8}+\dfrac{4t + 8}{3t + 1} $
Explanation: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(t - 8)(3t + 1)$ Multiply the first term by $\dfrac{3t + 1}{3t + 1}$ $ \begin{align*} \dfrac{t}{t - 8} \times \dfrac{3t + 1}{3t + 1} & = \dfrac{(t)(3t + 1)}{(t - 8)(3t + 1)} \\ & = \dfrac{3t^2 + t}{(t - 8)(3t + 1)}\end{align*} $ Multiply the second term by $\dfrac{t - 8}{t - 8}$ $ \begin{align*} \dfrac{4t + 8}{3t + 1} \times \dfrac{t - 8}{t - 8} & = \dfrac{(4t + 8)(t - 8)}{(3t + 1)(t - 8)} \\ & = \dfrac{4t^2 - 24t - 64}{(3t + 1)(t - 8)}\end{align*} $ Now we have: $ = \dfrac{3t^2 + t}{(t - 8)(3t + 1)} + \dfrac{4t^2 - 24t - 64}{(3t + 1)(t - 8)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{3t^2 + t + 4t^2 - 24t - 64}{(t - 8)(3t + 1)} $ $ = \dfrac{7t^2 - 23t - 64}{(t - 8)(3t + 1)}$ Expand the denominator: $ = \dfrac{7t^2 - 23t - 64}{3t^2 - 23t - 8}$